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أوجد المشتق العكسي لـ
∫11+sinxdx\int_{ }^{ }\frac{1}{1+\sin x}dx∫1+sinx1dx
sinx−tanx+c\sin x-\tan x+csinx−tanx+c
−secx+tanx+c-\sec x+\tan x+c−secx+tanx+c
secx−tanx+c\sec x-\tan x+csecx−tanx+c
أكمل هذا البيان
إذا كانت
f (x)=5x3 و g(x)=2x2
إذن
∫[f(x) -g(x)] dx = ___________.
54x4+ex+c\frac{5}{4}x^4+e^x+c45x4+ex+c
54x4−23x3+c\frac{5}{4}x^4-\frac{2}{3}x^3+c45x4−32x3+c
54x4+73x3+x+c\frac{5}{4}x^4+\frac{7}{3}x^3+x+c45x4+37x3+x+c
إذا كان
f (x)=Cos x و g(x)=x3
فإن
∫[f(x) - g(x)] dx = ___________.
x22+sinx+c\frac{x^2}{2}+\sin x+c2x2+sinx+c
sinx+14x4+c\sin x+\frac{1}{4}x^4+csinx+41x4+c
sinx−14x4+c\sin x-\frac{1}{4}x^4+csinx−41x4+c
∫xx2+2x+1dx\int_{ }^{ }\frac{x}{x^2+2x+1}dx∫x2+2x+1xdx
ln∣x+1∣+c\ln\left|x+1\right|+cln∣x+1∣+c
ln∣x+1∣+1x+1+c\ln\left|x+1\right|+\frac{1}{x+1}+cln∣x+1∣+x+11+c
−19x9+c-\frac{1}{9x^9}+c−9x91+c
قيّم التكامل
∫2x2x+1dx\int_{ }^{ }\frac{2x^2}{x+1}dx∫x+12x2dx
x2−2xln∣4x∣+cx^2-2x\ln\left|4x\right|+cx2−2xln∣4x∣+c
4ln∣x∣+c4\ln\left|x\right|+c4ln∣x∣+c
x2−2x+2ln∣x+1∣+cx^2-2x+2\ln\left|x+1\right|+cx2−2x+2ln∣x+1∣+c
∫x+1(x−2)3dx\int_{ }^{ }\frac{x+1}{\left(x-2\right)^3}dx∫(x−2)3x+1dx
12(x+2)4\frac{1}{2}\frac{\left(x+2\right)}{4}214(x+2)
1−2x2(x−2)2\frac{1-2x}{2\left(x-2\right)^2}2(x−2)21−2x +c
f (x)=2Cos x و g(x)=ex
إذن ∫[f(x) - g(x)] dx = ___________.
x22+ex+c\frac{x^2}{2}+e^x+c2x2+ex+c
2sinx−ex+c2\sin x-e^x+c2sinx−ex+c
2ex+x+c2e^x+x+c2ex+x+c
∫x+1x−2dx\int_{ }^{ }\frac{x+1}{x-2}dx∫x−2x+1dx
(x+3)ln∣x+1∣+c\left(x+3\right)\ln\left|x+1\right|+c(x+3)ln∣x+1∣+c
(x−3)ln∣x+1∣+c\left(x-3\right)\ln\left|x+1\right|+c(x−3)ln∣x+1∣+c
x+ln∣x−2∣+cx+\ln\left|x-2\right|+cx+ln∣x−2∣+c
f (x)=Sinx و g(x)=loglogx
2x2cosx−xloglogx+c2x^2\cos x-x\log\log x+c2x2cosx−xloglogx+c
−cosx+xloglogx−x+c-\cos x+x\log\log x-x+c−cosx+xloglogx−x+c
−cosx−xloglogx+x+c-\cos x-x\log\log x+x+c−cosx−xloglogx+x+c
∫1x2+4x+8dx\int_{ }^{ }\frac{1}{x^2+4x+8}dx∫x2+4x+81dx
12 x+24\frac{1}{2}\ \frac{x+2}{4}21 4x+2
12 x+22+c\frac{1}{2}\ \frac{x+2}{2}+c21 2x+2+c
إنتهى الإختبار.